So there is this quite well known Prisoner's dilemma where two parties can both defect and cooperate (and get points based on their decisions). In my presently used example I take it that cooperating player always gets $2$ points while defecting player gets $3$ points against cooperating opponent and $0$ points against defecting opponent.

So basically I have my $N^2$ imaginary players placed on a $N \times N$ periodic square lattice and every player (representing a single strategy) takes part in eight interactions (as described above) against adjacent elements. The sum of points received during eight mentioned interactions represents the success of a player.

Finally every player compares the points in one's neighborhood (i. e. himself and eight adjacent elements) and copies the strategy which was the most successful.

What is interesting is that system tends to come to a state where strategies do not change (it is obvious that if two consecutive iterations were identical then the same can be said about all further iterations) or in other words we reach the Nash equilibrium.

But what can one say about the sheer number of different states of Nash equilibrium? I can start with $2^{N^2}$ different states and computational experiments reveal that there is quite a few different states of equilibrium to be reached, but I am interested in purely mathematical point of view. What kind of evaluation, approximation, lower bound is possible? I would be thankful for your observations.